The orbital angular momentum for an electron revolving in an orbit is given by \sqrt{l\left ( l+1 \right )}.\frac{h}{2\pi }This momentum for an s- electron will be given by

 

  • Option 1)

    +\frac{1}{2}.\frac{h}{2\pi }

  • Option 2)

    zero

  • Option 3)

    \frac{h}{2\pi }

  • Option 4)

    \sqrt{2}\cdot \frac{h}{2\pi }

 

Answers (2)

As we learnt in 

Azimuthal Quantum Number(l) -

For a given value of n, l can have n values ranging from 0 to n – 1, that is, for a given value of n, the possible value of l are : l = 0, 1, 2, ....( n –1)

-

 

 The value of l ( azimuthal quantum number ) for s-electron is equal to zero.

Orbital\: angular \: momentum\: = \sqrt{l\left ( l+1 \right )\cdot \frac{h}{2\pi }}

Substituting the value of  l for s-electron = \sqrt{0\left ( 0+1 \right )}\cdot \frac{h}{2\pi }= 0


Option 1)

+\frac{1}{2}.\frac{h}{2\pi }

Incorrect option

Option 2)

zero

Correct Option

Option 3)

\frac{h}{2\pi }

Incorrect option

Option 4)

\sqrt{2}\cdot \frac{h}{2\pi }

Incorrect option

N neha

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